Ok. Got it. Take the expression you get for log(n) and factor all of the numbers into primes. Now if multiply everthing out and simplify, then, lo and behold, the result magically becomes log(N)/3. Where N was that number I was thinking of. Pretty painful - but I still can't think of a more direct way. Plenty of practice with the rules of logarithms there.

Ok, I got the answer after some very round about math. Right now I'm trying to find a way to do it with less work. I got the answer... n^3 is a two digit integer. If I figure out a way to do this more easily, I will throw you in the right direction.

Maybe we are taking this 'don't tell the answer thing' too seriously. I suspect that teleport knows the answer too. But this is getting to be sort of fun. So I'll say that the first letter of the last digit of the two digit number is 'f'. Your turn. Finding this with a lot less work would be a cool thing. But maybe somebody just cooked it to work this way.

If [itex]log_{4n}(40\sqrt{3})= log_{3n}(45)[/itex], then, letting a be that mutual value, [itex](4n)^a= 40\sqrt{3}[/itex] and [itex](3n)^a= 45[/itex]. Dividing one equation by the other [itex](4/3)^a= 8\sqrt{3}/9= 2^3/3^{3/2}[/itex]. That is, 2^{2a}= 2^3 and 3^a= 3^{3/2}: a= 3/2.